Problem: Let $a(x)=-18x^2-6x+12$, and $b(x)=3x^3+9x-1$. When dividing $a$ by $b$, we can find the unique quotient polynomial $q$ and remainder polynomial $r$ that satisfy the following equation: $\dfrac{a(x)}{b(x)}=q(x) + \dfrac{r(x)}{b(x)}$, where the degree of $r(x)$ is less than the degree of $b(x)$. What is the quotient, $q(x)$ ? $ q(x)=$ What is the remainder, $r(x)$ ? $r(x)=$
Solution: Note that $b(x)$ has a higher degree than $a(x)$. This implies that the quotient term, $q(x)$ is $0$. [Why?] We already know that $q(x)=0$. Since the degree of $a(x)$ is smaller than the degree of $b(x)$, the remainder $r(x)$ is simply $a(x)$. [How did we conclude this?] Therefore, $r(x)=-18x^2-6x+12$. To conclude, $q(x)=0$ $r(x)=-18x^2-6x+12$